Geometry Problem 1530. Post your solution in the comment box below.
Level: Mathematics Education, K-12 School, Honors Geometry, College.
Details: Click on the figure below.
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ABEF is a circle with AF as diameter.
ReplyDeleteSo < BFE = < BAE = 45 and so
< AFB = 90 - 45 - 20 = 25
Therefore < AEB = < AFB = 25
Sumith Peiris
Moratuwa
Sri Lanka
25°
ReplyDeleteABEF is a cyclic quadrilateral
(20+45)+(90+AEB)=180
AEB=25
x=25° : https://photos.app.goo.gl/d7LNrJchKUm4RTG66
ReplyDeleteABEF is a cyclic quadrilateral so ^(BEA)=^(BFA)
ReplyDeleteABCD is a square => ^(BCA)=^(ECF)=^(EFC)= 45
so ^(BFA)=^(BEA)=90-45-20=25
Angle ABF = Angle AEF = 90 degrees, hence Quadrilateral ABEF is cyclic with AF as diameter.
ReplyDeleteAngle BAE= Angle BFE =45 deg.
Angle AFB= 90 - 45 - 20 = 25 deg.
Since ABEF is cyclic Angle AEB = Angle AFB = 25 degrees.
25
ReplyDelete